\section{Basis Functions}

\subsection*{Problem 5}
Since we know that \[\tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \quad \textrm{and} \quad \sigma(x) = \frac{1}{1 + e^{-x}},\]
we can directly derive the desired equation:
\begin{eqnarray*}
2\sigma(2x)-1 &=& \frac{2}{1 + e^{-2x}} - 1 \\
              &=& \frac{2 - 1 - e^{-2x}}{1 + e^{-2x}} \\
              &=& \frac{1 - e^{-2x}}{1 + e^{-2x}} \\
              &=& \frac{e^x - e^{-x}}{e^x + e^{-x}}.
\end{eqnarray*}

For the second part, consider the following:
\begin{eqnarray*}
  y(x,w) &=& w_0 + \sum_jw_j\sigma\left(\frac{x-\mu_j}{s}\right) \\
         &=& w_0 + \sum_jw_j\left(\frac{\tanh\left(\frac{x-\mu_j}{2s}\right) + 1}{2} \right)\\
         &=& w_0 + \sum_j\frac{w_j}{2}\tanh\left(\frac{x-\mu_j}{2s}\right)  + \sum_jw_j.
\end{eqnarray*}

Hence, \[ u_0 = w_0 + \sum_jw_j, \qquad u_j = \frac{w_j}{2}.\]
